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C3 qu 4b)i) Ex 6F Heinemann’s book Query

Qu4.b)i)

Solve, in the interval 0 ≤ θ ≤ 360

sec (2θ - 15) = cosec 135

Now the obvious solution is simply:

cos (2θ - 15) = sin 135

2θ - 15 = cos-1 (sin 135)

2θ - 15 = 45, 315, 405, 675

θ = 30, 165, 210, 345

However, as the avid mathematician that I am, I felt like trying to do it without a calculator. Here is how I went about it: ( could have used addition rule, but this is a different method!)

sec (2θ - 15) = cosec 135

sec^{2} (2θ - 15) = cosec^{2} 135

1 + tan^{2} (2θ - 15) = 1 + cot^{2} 135

tan^{2} (2θ - 15) = cot^{2} 135

tan^{2} (2θ - 15) = (tan^{2} 135)^{-1}

I then worked out tan135 = -1 (using the tan graph, and the value tan45=1 – but a calculator is sufficient!!)

So:

tan^{2} (2θ - 15) = ([-1]^{2})-1

tan^{2} (2θ - 15) = 1

tan (2θ - 15) = ±1

Taking +1:

2θ - 15 = 45, 225, 405, 585

θ = 30, 120, 210, 300

And taking -1:

2θ - 15 = 135, 315, 495, 675

θ = 75, 165, 255, 345

SO: θ = 30, 75, 120, 165, 210, 255, 300, 345

Hence getting EXTRA answers, which don’t work if plugged back into first equation.

I suspect that this is to do with the stages marked with asterisks (see above) due to the ever changing signs. It could also have to do with the squaring done in the first step (perhaps?). Just wondering if someone could explain this for me please.

BTW:

Might be helpful to know that the incorrect answers (75, 120, 255, 300) are the negative of cosec135 when plugged into initial equation:

cosec135 = 2 ^ 0.5 = 1.414…

sec (2 x 75 - 15) = -1.414 = -2 ^ 0.5

sec (2 x 120 – 15) = -1.414 = -2 ^ 0.5

sec (2 x 255 – 15) = -1.414 = -2 ^ 0.5

sec (2 x 300 – 15) = -1.414 = -2 ^ 0.5

It might also have to do with the fact that i turned a cos solution into a tan one. And since cos is [360 - θ] while tan is [180 + θ] for diff solutions, it could affect the answers.

Sorry if anyone feels that this thread is a waste of time (due to me already getting the answers) but I believe in understanding maths to my fullest potential. . Thanks for replies.

Qu4.b)i)

Solve, in the interval 0 ≤ θ ≤ 360

sec (2θ - 15) = cosec 135

Now the obvious solution is simply:

cos (2θ - 15) = sin 135

2θ - 15 = cos-1 (sin 135)

2θ - 15 = 45, 315, 405, 675

θ = 30, 165, 210, 345

However, as the avid mathematician that I am, I felt like trying to do it without a calculator. Here is how I went about it: ( could have used addition rule, but this is a different method!)

sec (2θ - 15) = cosec 135

sec

1 + tan

tan

tan

I then worked out tan135 = -1 (using the tan graph, and the value tan45=1 – but a calculator is sufficient!!)

So:

tan

tan

tan (2θ - 15) = ±1

Taking +1:

2θ - 15 = 45, 225, 405, 585

θ = 30, 120, 210, 300

And taking -1:

2θ - 15 = 135, 315, 495, 675

θ = 75, 165, 255, 345

SO: θ = 30, 75, 120, 165, 210, 255, 300, 345

Hence getting EXTRA answers, which don’t work if plugged back into first equation.

I suspect that this is to do with the stages marked with asterisks (see above) due to the ever changing signs. It could also have to do with the squaring done in the first step (perhaps?). Just wondering if someone could explain this for me please.

BTW:

Might be helpful to know that the incorrect answers (75, 120, 255, 300) are the negative of cosec135 when plugged into initial equation:

cosec135 = 2 ^ 0.5 = 1.414…

sec (2 x 75 - 15) = -1.414 = -2 ^ 0.5

sec (2 x 120 – 15) = -1.414 = -2 ^ 0.5

sec (2 x 255 – 15) = -1.414 = -2 ^ 0.5

sec (2 x 300 – 15) = -1.414 = -2 ^ 0.5

It might also have to do with the fact that i turned a cos solution into a tan one. And since cos is [360 - θ] while tan is [180 + θ] for diff solutions, it could affect the answers.

Sorry if anyone feels that this thread is a waste of time (due to me already getting the answers) but I believe in understanding maths to my fullest potential. . Thanks for replies.

Probably because you squared out the sec and cosec, you put in the extra values....

ie, if you had 3=x and you squared it, it would give the same as if you squared -3=x, so when you get back to the original equation, you have twice as much considered and you end up with double the amount of answers to compensate for it, so watch out!

Also, as an additional point:

sec (2θ - 15) = cosec 135

is not always the same as saying

cos (2θ - 15) = sin 135

so be careful

ie, if you had 3=x and you squared it, it would give the same as if you squared -3=x, so when you get back to the original equation, you have twice as much considered and you end up with double the amount of answers to compensate for it, so watch out!

Also, as an additional point:

sec (2θ - 15) = cosec 135

is not always the same as saying

cos (2θ - 15) = sin 135

so be careful

Feria

Also, as an additional point:

sec (2θ - 15) = cosec 135

is not always the same as saying

cos (2θ - 15) = sin 135

so be careful

sec (2θ - 15) = cosec 135

is not always the same as saying

cos (2θ - 15) = sin 135

so be careful

Sorry but sec x = cosec y

is ALWAYS cos x = sin y.

Heres why:

sec x = cosec y

1/cosx = 1/siny

cross multiply

cos x = sin y

lol.

I was thinking it could be the squaring, but i am having second thoughts. this is because, yes we do square, but later on we square root anyway, take both positive and negative values.

Therefore, if x=3, or x=-3, x

thanks for the reply feria

Feria

Also, you have jumbled things about before squarerooting though, which is why wrongs and rights could have appeared in both

true. it still wouldnt explain why the incorrect answers give exactly -cosec135 instead of cosec135! i.e. the jumbling would then cause answers to be quite a bit different.

you have been solving two different eqns!

originally you had, say

cos(2x - phi) = K

ok, let y = cos(2x - phi) - K and solve for y = 0. 1st problem

now square the terms,

cos²(2x - phi) = K²

now, let y = cos²(2x - phi) - K²

y = (cos(2x - phi) - K)(cos(2x - phi) + K)

and solve for y = 0, 2nd problem

in the 2nd problem, when y = 0 then,

cos(2x - phi) - K = 0, which gives the solutions to the 1st problem

and

cos(2x - phi) + K = 0, which gives the additional solutions you found.

originally you had, say

cos(2x - phi) = K

ok, let y = cos(2x - phi) - K and solve for y = 0. 1st problem

now square the terms,

cos²(2x - phi) = K²

now, let y = cos²(2x - phi) - K²

y = (cos(2x - phi) - K)(cos(2x - phi) + K)

and solve for y = 0, 2nd problem

in the 2nd problem, when y = 0 then,

cos(2x - phi) - K = 0, which gives the solutions to the 1st problem

and

cos(2x - phi) + K = 0, which gives the additional solutions you found.

Fermat

you have been solving two different eqns!

originally you had, say

cos(2x - phi) = K

ok, let y = cos(2x - phi) - K and solve for y = 0. 1st problem

now square the terms,

cos²(2x - phi) = K²

now, let y = cos²(2x - phi) - K²

y = (cos(2x - phi) - K)(cos(2x - phi) + K)

and solve for y = 0, 2nd problem

in the 2nd problem, when y = 0 then,

cos(2x - phi) - K = 0, which gives the solutions to the 1st problem

and

cos(2x - phi) + K = 0, which gives the additional solutions you found.

originally you had, say

cos(2x - phi) = K

ok, let y = cos(2x - phi) - K and solve for y = 0. 1st problem

now square the terms,

cos²(2x - phi) = K²

now, let y = cos²(2x - phi) - K²

y = (cos(2x - phi) - K)(cos(2x - phi) + K)

and solve for y = 0, 2nd problem

in the 2nd problem, when y = 0 then,

cos(2x - phi) - K = 0, which gives the solutions to the 1st problem

and

cos(2x - phi) + K = 0, which gives the additional solutions you found.

thank you fermat for clearing that up. I thought that since i jumbled it about, the explanation would be longer, but it is very simple. Now i feel stupid i didnt get it in the first place without help!

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